NCERT Exemplars Electrostatic Potential & Capacitance - NCERT Exemplar

Multiple Choice Questions - I

Q.1. A capacitor of 4 µ F is connected as shown in the circuit (Fig. 2.1). The internal resistance of the battery is 0.5Ω . The amount of charge on the capacitor plates will be

(a) 0
(b) 4µ C
(c) 16µ C
(d) 8µ C

Ans. (d)

Solution.

Key concept: A capacitor behaves as a short circuit when uncharged (initially) and as an open circuit in steady DC (fully charged).

At steady state the capacitor acts as an open circuit in a DC network, so no current flows through the branch containing the capacitor and the 10 Ω resistor. Hence the potential difference across the 10 Ω resistor is zero, and the potential difference across the capacitor equals the potential between points A and B.

Current through the 2 Ω resistor (left branch) is given by

Current I = V/(R + r) = 5/(2 + 0.5) = 1 A.

Potential difference across 2 Ω resistor, V = I R = 1 × 2 = 2 V.

Thus potential difference across the capacitor is 2 V.

Charge on the capacitor is q = C V = 4 µF × 2 V = 8 µC.

Q.2. A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge
(a) Remains a constant because the electric field is uniform.
(b) Increases because the charge moves along the electric field.
(c) Decreases because the charge moves along the electric field.
(d) Decreases because the charge moves opposite to the electric field.

Ans. (c)

Solution.

Key concept: Electric potential decreases in the direction of the electric field. A positive charge released from rest in a uniform field moves in the direction of the field.

The electric field does positive work on the positive charge as it moves along the field.

Work done by the field is equal to the decrease in electric potential energy, so the electric potential energy of the positive charge decreases.

Q.3. Figure 2.2 shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

(a) The work done in Fig. (i) is the greatest.

(b) The work done in Fig. (ii) is least.

(c) The work done is the same in Fig. (i), Fig. (ii) and Fig. (iii).

(d) The work done in Fig. (iii) is greater than Fig. (ii)but equal to that in Fig. (i).

Ans. (c)

Solution.

Key concepts about equipotentials:

• The density of equipotential lines gives an idea of field magnitude: higher density → larger field strength.
• The electric field is perpendicular to equipotential surfaces or lines.
• Equipotential surfaces for a point charge are concentric spheres.
• For a uniform electric field, equipotentials are planes perpendicular to the field lines.
• A metallic surface is an equipotential surface.
• Equipotential surfaces do not cross.
• Work done in moving a charge along an equipotential is zero.

In all three figures the initial and final potentials at A and B are the same. For the same charge moved between the same potentials, the work done depends only on the potential difference, which is zero here. Therefore the work done is the same (zero) in all three cases.

Q.4. The electrostatic potential on the surface of a charged conducting sphere is 100V. Two statments are made in this regard:
S₁ : At any point inside the sphere, electric intensity is zero.
S₂ : At any point inside the sphere, the electrostatic potential is 100V.
Which of the following is a correct statement?
(a) S₁ is true but S₂ is false.
(b) Both S₁ & S₂ are false.
(c) S₁ is true, S₂ is also true and S₁ is the cause of S₂.
(d) S₁ is true, S₂ is also true but the statements are independant.

Ans. (c)

Solution.

The relation between electric field and potential is E = - dV/dr.

If E = 0 inside the conducting sphere, then dV/dr = 0, so V is constant inside the sphere. Therefore S1 is true and S2 is true; the zero field (S1) causes the uniform potential (S2).

Important points:

• Zero electric field does not imply zero potential. For example, inside a charged spherical shell E = 0 but V may be non-zero.
• Two equal and opposite charges can make potential zero at the midpoint while the electric field there is non-zero.

Q.5. Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately
(a) Spheres
(b) Planes
(c) Paraboloids
(d) Ellipsoids.

Ans. (a)

Solution.

At large distances the net collection of charges with non-zero total charge behaves like a point charge. Equipotential surfaces of a point charge are spherical surfaces, since potential depends only on radial distance:

V = q / (4πϵ0 r).

Q.6. A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d₁ and dielectric constant k₁ and the other has thickness d₂ and dielectric constant k₂ as shown in Fig. 2.3. This arrangement can be thought as a dielectric slab of thickness d (= d₁+d₂) and effective dielectric constant k. The k is

(a) 

(b) 

(c) 

(d) 

Ans. (c)

Solution.

The two dielectric regions between the plates act as two capacitors C1 and C2 in series.

Capacitance of first block: C1 = (ϵ0 k1 A)/d1.

Capacitance of second block: C2 = (ϵ0 k2 A)/d2.

Equivalent series capacitance C satisfies 1/C = 1/C1 + 1/C2.

Comparing expressions leads to the effective dielectric constant k given by

Multiple Choice Questions - Ii

Q.7. Consider a uniform electric field in the 

 direction. The potential is a constant

(a) In all space

(b) For any x for a given z

(c) For any y for a given z

(d) On the x-y plane for a given z

Ans. (b, c, d)

Solution.

The relation between field and potential is E = -∇V. A uniform electric field in the z-direction implies equipotential surfaces are planes parallel to the x-y plane. Hence for a fixed z, V is constant for all x and y; equivalently V is constant for any x for a given z, for any y for a given z, and on the x-y plane at that z.

Q.8. Equipotential surfaces
(a) Are closer in regions of large electric fields compared to regions of lower electric fields.
(b) Will be more crowded near sharp edges of a conductor.
(c) Will be more crowded near regions of large charge densities.
(d) Will always be equally spaced.

Ans. (a, b, c)

Solution.

Electric field magnitude is related to potential difference and separation between equipotentials. The electric field is approximately E ≈ ΔV/Δr; therefore smaller separation Δr (closer equipotentials) corresponds to larger E.

Sharp conductor edges and regions with high surface charge density have larger local E, so equipotentials are crowded there. Hence options (a), (b) and (c) are correct; (d) is false.

Q.9. The work done to move a charge along an equipotential from A to B

(a) Cannot be defined as 

(b) Must be defined as 

(c) Is zero

(d) Can have a non-zero value

Ans. (c)

Solution. 

On an equipotential surface V2 - V1 = 0.

Work done in moving charge q is W = q (V2 - V1) = 0.

Therefore the work done is zero. The expressions in options (a) and (b) involve specific numerical values for q (e.g., q = 1 C), so they cannot be generally asserted.

Q.10. In a region of constant potential
(a) The electric field is uniform
(b) The electric field is zero
(c) There can be no charge inside the region.
(d) The electric field shall necessarily change if a charge is placed outside the region.

Ans. (b, c)

Solution. 

Since E = -dV/dr, if V is constant then dV/dr = 0 and hence E = 0. Thus (b) is correct.

If there were charge within the region, the field would not be zero there, so (c) is also correct. Placing a charge outside need not necessarily change the field inside a closed equipotential region unless field lines penetrate; however, (d) is not a general necessity.

Q.11. In the circuit shown in Fig. 2.4. initially key K₁ is closed and key K₂ is open. Then K₁ is opened and K₂ is closed (order is important). [Take Q₁′ and Q₂′ as charges on C₁ and C₂ and V₁ and V₂ as voltage respectively.]

Then
(a) Charge on C₁ gets redistributed such that V₁ = V₂
(b) Charge on C₁ gets redistributed such that Q₁′ = Q₂,′
(c) Charge on C₁ gets redistributed such that C₁V₁ + C₂V₂ = C₁ E
(d) Charge on C₁ gets redistributed such that Q₁′ + Q₂′ = Q

Ans. (a, d)

Solution. 

Initially K1 closed charges C1 by the battery; C2 is uncharged. After opening K1 and closing K2, C1 and C2 are connected in parallel so they share charge until their potentials equalise, hence V1 = V2.

Charge is conserved: the total charge initially on C1 equals the sum of charges on C1 and C2 after redistribution, so Q1′ + Q2′ = Q.

Q.12. If a conductor has a potential V ≠ 0 and there are no charges anywhere else outside, then
(a) There must be charges on the surface or inside itself
(b) There cannot be any charge in the body of the conductor
(c) There must be charges only on the surface
(d) There must be charges inside the surface

Ans. (a, b)

Solution. 

The potential of an isolated body must arise from its own charge if there are no external charges, so the body must carry charge (option a).

In electrostatic equilibrium charges reside on the outer surface of a conductor; therefore there cannot be charge in the conducting material (option b). Charge can, however, exist on inner cavity surfaces if charges are enclosed in cavities.

Q.13. A parallel plate capacitor is connected to a battery as shown in Fig. 2.5. Consider two situations:

A: Key K is kept closed and plates of capacitors are moved apart using insulating handle.

B: Key K is opened and plates of capacitors are moved apart using insulating handle.

Choose the correct option(s).
(a) In A : Q remains same but C changes.
(b) In B : V remains same but C changes.
(c) In A : V remains same and hence Q changes.
(d) In B : Q remains same and hence V changes.

Ans. (c, d)

Solution. 

When the capacitor remains connected to the battery (Case A), the battery keeps V constant; changing plate separation changes C, so Q = C V changes. Thus (c) is correct.

When the capacitor is isolated (battery disconnected, Case B), the charge Q is constant; changing plate separation changes C so V = Q/C changes. Thus (d) is correct.

Very Short Answer Type Questions

Q.14. Consider two conducting spheres of radii R₁ and R₂ with R₁ > R₂. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Ans. 

Let the charges be Q1 and Q2 on spheres of radii R1 and R2 respectively. For isolated spheres, V = Q/(4πϵ0 R).

Since V1 = V2, Q1/R1 = Q2/R2, therefore Q1/Q2 = R1/R2.

Surface charge density σ = Q / (4π R^2). So σ1/σ2 = (Q1/R1^2) / (Q2/R2^2) = (R1/R2) × (R2^2/R1^2) = R2/R1.

Because R1 > R2, σ1 < σ2. Thus the smaller sphere has a larger surface charge density than the larger sphere.

Q.15. Do free electrons travel to region of higher potential or lower potential?

Ans. 

Free electrons (negative charges) experience force opposite to the electric field direction.

Electric field points from higher potential to lower potential, so electrons move opposite to the field, i.e., from lower potential to higher potential.

Q.16. Can there be a potential difference between two adjacent conductors carrying the same charge?

Ans. 

Yes. Capacitance (and hence potential for given charge) depends on geometry. Two adjacent conductors of different sizes or shapes but carrying the same charge may have different potentials since V = Q/C and C depends on geometry.

Q.17. Can the potential function have a maximum or minimum in free space?

Ans. 

In electrostatics (in charge-free regions), the potential satisfies Laplace's equation and therefore cannot have a local maximum or minimum in free space; extremes occur only at boundaries or where charges are present.

Q.18. A test charge q is made to move in the electric field of a point charge Q along two different closed paths (Fig. 2.6).First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?

Ans. 

Work done is zero for both closed paths because the electrostatic field is conservative; the work done around any closed loop is zero.

Short Answer Type Questions

Q.19. Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

Ans. 

Assume a closed surface is equipotential but the potential just inside varies. Then dV/dr ≠ 0 inside, so E = -dV/dr ≠ 0 and field lines would cross the surface. Field lines cannot lie along an equipotential surface, so they must originate from or terminate on charges inside; this contradicts the assumption of no internal charges. Hence the potential must be constant throughout the volume enclosed by the closed equipotential surface; the interior is an equipotential volume.

Q.20. A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

Ans. 

Let initial dielectric constant be K (>1). Capacitance with dielectric C = K ϵ0 A/d.

When dielectric is removed (K → 1):

• Capacitance C decreases.
• If battery disconnected, charge q remains constant (conservation of charge).
• Energy U = q^2 / (2 C) increases because C decreases.
• Voltage V = q / C increases because C decreases.
• Electric field E = V / d increases since V increases.

Q.21. Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Ans. 

Potential decreases along electric field lines. Starting at the charged conductor and moving along a field line toward the uncharged conductor the potential decreases; continuing from the uncharged conductor to infinity the potential further decreases to the potential at infinity (zero). Thus the uncharged conductor, once induced, must take a potential intermediate between that of the charged conductor and infinity.

Q.22. Calculate potential energy of a point charge -q placed along the axis due to a charge +Q uniformly distributed along a ring of radius R. Sketch P.E. as a function of axial distance z from the centre of the ring. Looking at graph, can you see what would happen if -q is displaced slightly from the centre of the ring (along the axis)?

Ans. 

Potential at axial distance z due to ring of total charge Q and radius a (or R):

Potential at axial point P is V(z) = (1 / (4πϵ0)) · Q / sqrt(z^2 + R^2).

Potential energy of -q at that point is U(z) = q V(z) with q negative, so

The U(z) curve is symmetric about z = 0 and has a maximum or minimum depending on sign; a negative charge placed at the centre will be in stable or unstable equilibrium depending on curvature. For a negative test charge at the centre of a positively charged ring, small axial displacements lead to restoring forces, producing oscillations (small oscillations about z = 0).

Q.23. Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.

Ans. 

Consider a ring of radius R with total charge Q uniformly distributed. For a point P on axis at distance z from centre O, each element dq contributes dV = (1/(4πϵ0)) · dq / r where r = sqrt(z^2 + R^2).

Integrating over the ring (dq sums to Q) gives

Thus V_p = (1/(4πϵ0)) · Q / sqrt(z^2 + R^2).

Long Answer Type Questions

Q.24. Find the equation of the equipotentials for an infinite cylinder of radius r₀, carrying charge of linear density λ.

Ans. 

Use a cylindrical Gaussian surface of radius r (r > r0) and length l. Electric field is radial and uniform on the curved surface.

By Gauss's law:

∮ E · dA = E (2π r l) = q_enclosed / ϵ0 = λ l / ϵ0.

So E(r) = λ / (2πϵ0 r).

Potential difference between r0 and r is

V(r) - V(r0) = - ∫_{r0}^{r} E(r') dr' = - (λ / (2πϵ0)) ∫_{r0}^{r} (1/r') dr' = - (λ / (2πϵ0)) ln(r / r0).

Thus equipotential surfaces are coaxial cylinders with radii related by ln(r / r0) = constant, so r = r0 exp(-constant · (2πϵ0)/λ). In short, equipotentials are cylindrical surfaces concentric with the charged cylinder.

Q.25. Two point charges of magnitude + q and - q are placed at (-d/2, 0,0) and (d/2, 0,0), respectively. Find the equation of the equipoential surface where the potential is zero.

Ans. 

Potential at a point (x,y,z) due to +q and -q is

V = (1/(4πϵ0)) [ q/ r1 - q/ r2 ] where r1 and r2 are distances to the +q and -q respectively.

Set V = 0 ⇒ 1/r1 = 1/r2 ⇒ r1 = r2.

The locus of points equidistant from the two charges is the plane perpendicular to the line joining them at the midpoint: the plane x = 0 (the Y-Z plane).

Q.26. A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where α = 2V^-1. A similar capacitor with no dielectric is charged to U₀ = 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Ans. 

Let capacitance of the empty capacitor be C. Initial charge on it is q0 = C U0 = 78 C.

When connected, let final common voltage be U.

Capacitance of dielectric-filled capacitor is C2 = ε C = α U C (because ε = α U), so its charge q2 = C2 U = α U C · U = α C U^2.

Charge conservation: 78 C = C U + α C U^2.

Divide by C: 78 = U + α U^2 with α = 2 V^-1.

Thus 78 = U + 2 U^2 ⇒ 2 U^2 + U - 78 = 0.

Solving positive root yields U = 6 V.

Q.27. A capacitor is made of two circular plates of radius R each, separated by a distance d<<R. The capacitor is connected to a constant voltage. A thin conducting disc of radius r<<R and thickness t<<r is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is m.

Ans. 

Plates A and B are horizontal, separation d << R, and the capacitor is maintained at potential V.

Electric field magnitude between plates E ≈ V/d.

When the conducting disc is placed on the bottom plate, it acquires induced charge q' from the plate.

By Gauss's law for a small pillbox around disc area:

Induced charge on disc q' = ϵ0 E × area = ϵ0 (V/d) × π r^2.

Electrostatic upward force on disc due to plate is F = q' E = q' (V/d).

Substitute q' to get F in terms of V:

Equate to weight mg to find minimum V such that F ≥ mg:

Solving yields the required minimum voltage V_min (expression shown in images).

Q.28. (a) In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3) e] and two down quarks [charges -(1/3) e]. Assume that they have a triangle configuration with side length of the order of 10^-15 m. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.
(b) Repeat above exercise for a proton which is made of two up and one down quark.

Ans. 

Use Coulomb potential energy between point charges. For neutron (one up and two down), compute pairwise interactions using charges q_u = +2e/3 and q_d = -e/3, separation r ≈ 10^-15 m.

Summing pairwise Coulomb energies (with appropriate signs) gives total electrostatic potential energy U ≈ -0.48 MeV (attractive net), which is tiny compared to neutron rest energy 939 MeV.

Thus electrostatic contribution is ≪ rest mass energy (ratio ≈ 5.11 × 10^-4).

For proton (two up, one down) repeat similar calculation; images show details and numerical estimate.

Q.29. Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density σ. They are brought in contact and separated. What will be new surface charge densities on them?

Ans. 

Initial charges: for sphere of radius R, q1 = σ × 4π R^2. For radius 2R, q2 = σ × 4π (2R)^2 = σ × 16π R^2.

Total initial charge Q_total = q1 + q2 = σ 4π R^2 (1 + 4) = 5 σ 4π R^2.

When spheres touched, charge redistributes so potentials equal: q1_final / R = q2_final / (2R) ⇒ q2_final = 2 q1_final.

Conservation: q1_final + q2_final = Q_total ⇒ q1_final + 2 q1_final = 5 q1_initial ⇒ 3 q1_final = 5 q1_initial.

Solving gives q1_final = (5/3) q1_initial and q2_final = (10/3) q1_initial.

New surface charge densities are σ1' = q1_final / (4π R^2) and σ2' = q2_final / (4π (2R)^2). Compute explicitly with numbers shown in images.

Q.30. In the circuit shown in Fig. 2.7, initially K₁ is closed and K₂ is open. What are the charges on each capacitors.

Then K₁ was opened and K₂ was closed (order is important), What will be the charge on each capacitor now? [C = 1µF]

Ans.

When K1 closed and K2 open, C1 and C2 are charged by the 9 V battery. Let potentials across C1 and C2 be V1 and V2. Then V1 + V2 = 9 V.

Using capacitance ratios (from circuit geometry shown in image) one finds V2 = 2 V1.

Solving V1 + 2 V1 = 9 ⇒ V1 = 3 V, V2 = 6 V.

With C = 1 µF, q1 = C1 V1 = 6C × 3 = 18 µC and q2 = C2 V2 = 3C × 6 = 18 µC, hence q1 = q2 = 18 µC.

Now K1 opened and K2 closed: q2 redistributes between C2 and C3 (which are in parallel) but total charge on that parallel pair remains 18 µC. If C2 and C3 are equal (3 µF each as in diagram), the voltage becomes V = 18 µC / (6 µF) = 3 V. Thus potentials on C2 and C3 are each 3 V.

Q.31. Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.

Ans. 

Consider a point P on the axis at distance x from disc centre. Take a concentric ring element of radius r and width dr. Charge on ring dq = σ · (2π r dr).

Potential due to ring at P: dV = (1/(4πϵ0)) · dq / sqrt(r^2 + x^2).

Integrate r from 0 to R:

Carrying out the integral yields

Which after simplification gives V = (1/(4πϵ0)) · (2πσ) [ sqrt(R^2 + x^2) - x ]. Rewriting σ in terms of Q yields the expression shown in images.

Q.32. Two charges q₁ and q₂ are placed at (0, 0, d) and (0, 0, -d) respectively. Find locus of points where the potential a zero.

Ans. 

Let potential at point P(x,y,z) be zero:

V = (1/(4πϵ0)) [ q1 / r1 + q2 / r2 ] = 0.

Using algebraic manipulation (componendo and dividendo, as shown in images) leads to the equation of a sphere (or other quadric depending on signs and magnitudes of q1 and q2). The images show the full algebraic steps arriving at the sphere equation for the specific case when charges have opposite signs or equal magnitudes as required.

Q.33. Two charges -q each are separated by distance 2d. A third charge + q  is kept at mid point O. Find potential energy of + q as a function of small distance x from O due to -q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.

Ans. 

Place +q at a small displacement x from the midpoint toward one negative charge. Distances to the two -q charges become (d - x) and (d + x).

Potential energy U(x) = (1/(4πϵ0)) [ -q^2/(d - x) - q^2/(d + x) ].

Expand for small x (Taylor expansion) to obtain U(x) ≈ constant + (q^2/(4πϵ0)) · (2 x^2 / d^3) × (+ sign or negative sign depending on algebra). The quadratic term has the sign indicating whether U has a minimum or maximum at x = 0.

Calculation (shown in images) shows that the potential energy has a maximum at x = 0, so small displacements lower the potential energy - this corresponds to an unstable equilibrium at O.